[海量数据处理]用2-Bitmap找出数组中不重复的整数

题目来自: /article/1360519.html

5. 在2.5亿个整数中找出不重复的整数，内存不足以容纳这2.5亿个整数。

#include<stdio.h>
#include<memory.h>
//用char数组存储2-Bitmap,不用考虑大小端内存的问题
unsigned char flags[1000]; //数组大小自定义

unsigned get_val(int idx)
{
int i = idx/4;
int j = idx%4;
unsigned ret = (flags[i]&(0x3<<(2*j)))>>(2*j);
return ret;
}

unsigned set_val(int idx, unsigned int val)
{
int i = idx/4;
int j = idx%4;
unsigned tmp = (flags[i]&~((0x3<<(2*j))&0xff)) | (((val%4)<<(2*j))&0xff);
flags[i] = tmp;
return 0;
}
unsigned add_one(int idx)
{
if (get_val(idx)>=2) {
return 1;
}
else  {
set_val(idx, get_val(idx)+1);
return 0;
}
}

//只测试非负数的情况;
//假如考虑负数的话,需增加一个2-Bitmap数组.
int a[]={1, 3, 5, 7, 9, 1, 3, 5, 7, 1, 3, 5,1, 3, 1,10,2,4,6,8,0};

int main()
{
int i;
memset(flags, 0, sizeof(flags));

printf("原数组为:");
for(i=0;i < sizeof(a)/sizeof(int); ++i)  {
printf("%d  ", a[i]);
add_one(a[i]);
}
printf("\r\n");

printf("只出现过一次的数:");
for(i=0;i < 100; ++i)  {
if(get_val(i) == 1)
printf("%d  ", i);
}
printf("\r\n");

return 0;
}

除了用2-Bitmap来计数标记以外,也可以用两个1-Bitmap来实现(如果考虑正负数的情况,就四个1-Bitmap)

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